Question Number 186911 by Spillover last updated on 11/Feb/23
![∫_0 ^∞ (1/(1+a^x +a^(x/2) ))dx = (1/(ln a))[ln 3−(π/(3(√3)))]](Q186911.png)
$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{a}^{{x}} +{a}^{\frac{{x}}{\mathrm{2}}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\left[\mathrm{ln}\:\mathrm{3}−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\right] \\ $$
Answered by witcher3 last updated on 13/Feb/23
![a^(x/2) =t,a>1 x=2((ln(t))/(ln(a))) dx=(2/(ln(a)t))dt ⇔(2/(ln(a)))∫_1 ^∞ (dt/(t(1+t+t^2 ))) (2/(ln(a)))∫_0 ^1 y(dy/(1+y+y^2 )) =(2/(ln(a)))∫_0 ^1 ((y+(1/2))/(1+y+y^2 ))−(1/(ln(a)))∫_0 ^1 (dy/((y+(1/2))^2 +(3/4))) =(2/(ln(a))).(1/2)[ln(1+y+y^2 )]_0 ^1 −(1/(ln(a)))∫_0 ^1 .(4/3)(dy/(1+(((2y)/( (√3)))+(1/( (√3)))))) (1/(ln(a)))[ln(3)−(2/( (√3)))tan^(−1) ((√3))+(2/( (√3)))tan^(−1) ((1/( (√3))))) =(1/(ln(a)))[ln(3)−(π/( 3(√3)))]](Q187071.png)
$$\mathrm{a}^{\frac{\mathrm{x}}{\mathrm{2}}} =\mathrm{t},\mathrm{a}>\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{2}\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{ln}\left(\mathrm{a}\right)} \\ $$$$\mathrm{dx}=\frac{\mathrm{2}}{\mathrm{ln}\left(\mathrm{a}\right)\mathrm{t}}\mathrm{dt} \\ $$$$\Leftrightarrow\frac{\mathrm{2}}{\mathrm{ln}\left(\mathrm{a}\right)}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{t}\left(\mathrm{1}+\mathrm{t}+\mathrm{t}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{2}}{\mathrm{ln}\left(\mathrm{a}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{y}\frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}+\mathrm{y}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{ln}\left(\mathrm{a}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{y}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\mathrm{y}+\mathrm{y}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{a}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dy}}{\left(\mathrm{y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{ln}\left(\mathrm{a}\right)}.\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{y}+\mathrm{y}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{a}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} .\frac{\mathrm{4}}{\mathrm{3}}\frac{\mathrm{dy}}{\mathrm{1}+\left(\frac{\mathrm{2y}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{a}\right)}\left[\mathrm{ln}\left(\mathrm{3}\right)−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{a}\right)}\left[\mathrm{ln}\left(\mathrm{3}\right)−\frac{\pi}{\:\mathrm{3}\sqrt{\mathrm{3}}}\right] \\ $$